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When we expand this out, we get four terms, and not two terms as needed. So $L \otimes L$ is not a Lie algebra, and we can't ask $\Delta$ to be a Lie algebra map. But

Edit: There seems to be some related notions. One could ask that $\Delta: L \rightarrow L \otimes L$ is an $L$-module, and we can ask that $\Delta$ be a map of Lie modules. This is what is asked for in a Lie bialgebra. The analogous definition thing with associative algebras is an open (non-commutative) Frobenius algebra. This is an algebra $A$, which consists of a commutative A$ with multiplication $m$ and cocommutative comultiplication $\Delta: A \rightarrow A \otimes A$, such that $\Delta$ is a map of $A$ bimodulesmodules. That is,

One could also ask that $$\Delta(a\cdot b) = a_{(1)} \Delta: L \otimes rightarrow L \cdot a_{(2)} b + otimes L$ be a \cdot b_{(1)} derivation of $L$ with values in the bimodule $L \otimes b_{(2)}L$. $$

Added: Maybe it would be more accurate to say This is the cocycle condition. The analogous thing for associative algebras is called an infinitesimal bialgebra. Aguiar discusses these objects in the paper "On the associative analog of Lie bialgebras." The idea of symmetrizing the associative product of an infinitesimal bialgebra is analogous to obtain a non-commutative Frobenius Lie bialgebra is discussed (this process does not always yield a Lie bialgebra).

In "Skein quantization of Poisson algebras of loops on surfaces," Turaev describes a Lie bialgebra. He also discusses a Hopf algebra which quantizes the Lie bialgebra. That is, an associative Since a Hopf algebra $A$, is a bialgebra with coassociative comultiplication $\Delta$, where $\Delta$ an antipode, maybe there is a map of $A$-bimodules. connection to be said between Lie bialgebras and bialgebras.

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Lie bialgebras and bialgebras cannot be analogous. The tensor product of two Lie algebras is not a Lie algebra. We can check this, let $L_1$ and $L_2$ be two Lie algebras, so for $a,b,c \in L_1$ and $x,y,z \in L_2$ we have $$[a,[b,c]] = [[a,b], c] + (-1)^{|a||b|} [b, [a,c]] $$ $$[x, [y,z]] = [[x,y], z] + (-1)^{|x||y|} [y, [x,z]].$$ We use this version of the Jacobi identity-that $[,]$ is a derivation of itself- to keep track of signs, although it won't be that important.

So we can try to define a Lie bracket on $L_1 \otimes L_2$ by $[a\otimes b , x \otimes y]= [a,b] \otimes [x,y].$ But because the Jacobi identity has three terms (and not two like in the associative identity), we get

$$[a \otimes x, [b \otimes y, c \otimes z]] = [a \otimes x , [b,c] \otimes [y,z]] $$ $$= [a, [b,c] \otimes [x,[y,z]] $$ $$= \left([[a,b], c] + (-1)^{|a||b|} [b, [a,c]] \right) \otimes \left( [[x,y], z] + (-1)^{|x||y|} [y, [x,z]] \right).$$

When we expand this out, we get four terms, and not two terms as needed.

So $L \otimes L$ is not a Lie algebra, and we can't ask $\Delta$ to be a Lie algebra map. But $L \otimes L$ is an $L$-module, and we can ask that $\Delta$ be a map of Lie modules. This is what is asked for in a Lie bialgebra. The analogous definition is an open Frobenius algebra $A$, which consists of a commutative multiplication $m$ and cocommutative comultiplication $\Delta: A \rightarrow A \otimes A$ such that $\Delta$ is a map of $A$ bimodules. That is, $$\Delta(a\cdot b) = a_{(1)} \otimes \cdot a_{(2)} b + a \cdot b_{(1)} \otimes b_{(2)}. $$

Added: Maybe it would be more accurate to say the Lie bialgebra is analogous to a non-commutative Frobenius algebra. That is, an associative algebra $A$, with coassociative comultiplication $\Delta$, where $\Delta$ is a map of $A$-bimodules. These algebras are called infinitesimal bialgebras. Aguiar writes about them in "On the assocaitive analogs of Lie bialgebras." Apparently, you don't always get a Lie bialgebra by symmetrizing the multiplication and comultiplication.
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Lie bialgebras and bialgebras cannot be analogous. The tensor product of two Lie algebras is not a Lie algebra. We can check this, let $L_1$ and $L_2$ be two Lie algebras, so for $a,b,c \in L_1$ and $x,y,z \in L_2$ we have $$[a,[b,c]] = [[a,b], c] + (-1)^{|a||b|} [b, [a,c]] $$ $$[x, [y,z]] = [[x,y], z] + (-1)^{|x||y|} [y, [x,z]].$$ We use this version of the Jacobi identity-that $[,]$ is a derivation of itself- to keep track of signs, although it won't be that important.

So we can try to define a Lie bracket on $L_1 \otimes L_2$ by $[a\otimes b , x \otimes y]= [a,b] \otimes [x,y].$ But because the Jacobi identity has three terms (and not two like in the associative identity), we get

$$[a \otimes x, [b \otimes y, c \otimes z]] = [a \otimes x , [b,c] \otimes [y,z]] $$ $$= [a, [b,c] \otimes [x,[y,z]] $$ $$= \left([[a,b], c] + (-1)^{|a||b|} [b, [a,c]] \right) \otimes \left( [[x,y], z] + (-1)^{|x||y|} [y, [x,z]] \right).$$

When we expand this out, we get four terms, and not two terms as needed.

So $L \otimes L$ is not a Lie algebra, and we can't ask $\Delta$ to be a Lie algebra map. But $L \otimes L$ is an $L$-module, and we can ask that $\Delta$ be a map of Lie modules. This is what is asked for in a Lie bialgebra. The analogous definition is an open Frobenius algebra $A$, which consists of a commutative multiplication $m$ and cocommutative comultiplication $\Delta: A \rightarrow A \otimes A$ such that $\Delta$ is a map of $A$ bimodules. That is, $$\Delta(a\cdot b) = a_{(1)} \otimes \cdot a_{(2)} b + a \cdot b_{(1)} \otimes b_{(2)}. $$

Added: Maybe it would be more accurate to say the Lie bialgebra is analogous to a non-commutative Frobenius algebra. That is, an associative algebra $A$, with coassociative comultiplication $\Delta$, where $\Delta$ is a map of $A$-bimodules. These algebras are called infinitesimal bialgebras. Aguiar writes about them in "On the assocaitive analogs of Lie bialgebras." Apparently, you don't always get a Lie bialgebra by symmetrizing the multiplication and comultiplication.
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